2. Basics
Inverters are DC to AC converters
We can use inverters to generate
• A dc supply
• Single-phase AC supply
• Three-phase AC supply
from a single dc source.
The basic building block is the
inverter ‘leg’.
An inverter leg is shown. Vdc is
the input, Vout the output.
3. Inverter switching
s
on
dc
out
T
t
V
V ,
1
• T1 and T2 are NEVER turned on
together. Why?
• T1 and T2 are switched using PWM in a
complementary manner (T2 ON, T1 OFF)
• Vout is then a switched waveform, just
like the basic step-down converter earlier.
5. Current Paths
• Two switches with freewheel
diodes provides uni-directional
voltage and bi-directional
current control.
• Only when T1 is ON is energy
supplied from the source.
• When T2 ON, a zero voltage
loop is applied.
• With positive current flow →
Current path if T2 ON, or T1 and T2 OFF
Current path if T1 ON
6. Current Paths
• When T1 is ON (or T1 and T2
OFF) energy has to be
absorbed by the source.
• When T2 ON, a zero voltage
loop is applied.
• With negative current flow →
Current path if T2 ON
Current path if T1 ON, or T1 and T2 OFF
7. Bridge Leg V-I graph
The basic bridge leg can operate in two quadrants of the VI graph.
V
I
8. Average Output Voltage
A single inverter leg produces an average output voltage:
Define a duty cycle or modulation index
Hence
m must be between 0 and 1.
We can make m vary in time therefore we can produce any voltage and any
frequency we desire (within the bounds fixed by the switching
frequency and Vdc).
s
on
dc
out
T
t
V
V ,
1
s
on
T
t
m ,
1
dc
out mV
V
9. Switching Frequency
• Switch frequency (1/Ts) of the pulse-width modulated (PWM) signal is
usually chosen as high as possible to reduce current ripple in the load.
• Max switching frequency is limited by losses and the ability to manage
those device losses (remember lecture 2?)
• In low power circuits, switching frequency can be as high as ~1 MHz
• High power circuits (say >500kW) may use frequencies of 1kHz or less.
11. Single-phase H-bridge
• From previous discussion on inverter legs
• So the average output voltage
applied to the load
• For a sinusoidal output (ma-mb) must vary sinusoidally.
dc
b
bo
dc
a
ao
V
m
V
V
m
V
)
1
(
dc
b
a
bo
ao
load
V
m
m
V
V
V
)
3
(
sin
2
1
)
2
(
sin
2
1
t
m
m
t
m
m
b
a
12. Single-phase H-bridge
• The modulation indices of both inverter legs vary sinusoidally in time with a
modulation depth, m (0<m<0.5) and an offset. If we apply (2) and (3) to (1)
we get
• That is, the modulation depth, m, sets the magnitude of the ac output
voltage and ωt sets the frequency.
• Notice that the dc offset in the modulation indices is co-phasal and does
not appear in the output voltage.
• We can control the magnitude and frequency.
t
mV
V
t
m
t
m
V
dc
dc
load
sin
2
sin
2
1
sin
2
1
13. H-Bridge V-I graph
• The H-bridge can operate in all four quadrants of the VI graph.
• It can generate both polarities of voltage and control both polarities of
current.
V
I
14. PWM Generation
• Modulation indices of each
leg are compared with a
triangular carrier
waveform.
• Intersects define the turn-
on and turn-off instant of
each bridge leg.
• With this scheme load sees
two output voltage pulses
per switching cycle.
• Harmonic spectrum of the
applied voltage has
components around
multiples of the switching
frequency.
Ts
t1
t2
(t1-t2)/2
Ts/2
m1
m2
Vload
0
Leg 1 output
Leg 2 output
Carrier waveform
16. Three-phase inverters
Inserting modulation indices into (1-3) gives:
Three-phase output voltages
• The circuit is a pulse-width modulated voltage source inverter
(VSI).
2
sin
3
6
5
sin
3
6
sin
3
t
mV
V
t
mV
V
t
mV
V
dc
ca
dc
bc
dc
ab
17. Six-step Operation
• The previous section looked at pulse-width modulated VSIs.
• PWM VSIs can be used at all but very high power drives.
• For high-power drives, often the switches are turned ON and OFF once
during one fundamental cycle rather than many 100s of times with PWM.
• The output voltage waveform is then ‘square wave’.
D
-
Vd
+
o
1
4
S
S
D4
S
S
A B
D
S
6
1
3
S
6
D 3
C
D
D 2
2
5
5
0
S 6 on
S 1 on
S 6 on
_
A B =
V
V
V
V
AO BO
CO
V
0 60
BO
VAO
Vd
S 2 on
S 5 on
240
S 4 on
120
S 3 on
300
360
1 /
2V
d
S 2 on
1 /
2V
d S 5 on
S 3 on
1 /
2V
d
S 1 on
S 4 on
(
d )
t
t
(c )
(
b )
t
t
(
a )
18. Six-step Line Voltages
• Line voltages are
stepped
• Fourier analysis of
output voltages gives
• and line voltages
• and phase voltage
0
S 6 on
S 2
sw itch i
ng
sequen ce
S 2
S 1
BC
_
C A
_
V
VCO
V
VBO
=
V
AO
=
VCO
S 5
S 3 S 4 S 1
S 6
S 1 on
S 6 on
_
A B =
V
V
V
V
AO BO
CO
V
0 60
BO
Vd
S 2 on
S 5 on
240
S 4 on
120
S 3 on
300
360
1 /
2V
d
S 1
S 3 S 4 S 5 S 6
t
S 3
S 2
t
(
g )
(f)
(
e )
t
S 2 on
1 /
2V
d S 5 on
S 3 on
1 /
2V
d
S 1 on
S 4 on
(
d )
t
t
(c )
(
b )
t
t
(
a )
......
t
5
sin
5
1
t
3
sin
3
1
t
sin
2
V
.
4
V d
AO
.......
t
11
sin
11
1
t
7
sin
7
1
-
t
5
sin
5
1
-
t
sin
V
3
2
V d
AB
......
t
7
sin
7
1
t
5
sin
5
1
t
sin
V
2 d
19. Six-step Inverter Currents
• Inverter currents are obviously
non-sinusoidal. (Note: this load is
inductive)
• Result from the harmonic voltages
in the output line voltage.
• Harmonic currents causes
additional loss components.
• And also torque ripple if the load
is a machine.
deg rees
i
CA
i
A
AB
CA
- i
i
=
i
1
i
1
i
2
2
ii
1
i
120
i
i
AB
i
BC
i
1
0
V
AB
2
2
i
i
1
i
i
1
i
2
1
2
i
1
i
1
i
180 300
360
t
t
2
i
2
i
t
t
t
